Handbook of Electrical Engineering

426 HANDBOOK OF ELECTRICAL ENGINEERING

The combined admittance of the rotor and magnetising impedances become,

Ym 2 =

−j

nXm

+

R 2

Sn

−jnX 2

(

R 2

sn

) 2

+n^2 X 22

= 0. 01211 −j 1 .8650 ohms

Zm 2 =

1

Ym 2

= 0. 00348 +j 0 .5362 ohms

Add the stator impedance.

Z 1 m 2 =R 1 +jnX 1 +Zm 2 = 0. 00878 +j 0 .7711 ohms

All the harmonics of the supply voltage have zero phase shift (except the triplens which are
anti-phase). The fifth harmonic supply voltageV 5 nis 67. 77 /

√

2 volts (rms). Supply this voltage to
the circuit. The supply current is,

I 15 =

V 5

Z 1 m 2

=

67. 77 +j 0. 0

√

2 ( 0. 00878 +j 0. 7711 )

= 0. 7075 −j 62 .1377 amps

The volt-drop across the stator impedance is,

V 1 m 5 =( 0. 0053 +j 0. 2350 )( 0. 7075 −j 62. 1377 )

= 14. 606 −j 0 .163 volts

The air-gap voltageVm 5 becomes,

Vm 5 =V 15 −V 1 m 5

= 47. 921 − 14. 606 +j 0. 163

= 33. 315 +j 0 .163 volts

The magnetising currentIm 5 is,

Im 5 =

Vm 5

jnXm

=

33. 315 +j 0. 163

0. 0 +j 14. 665

= 0. 0111 −j 2 .272 amps

Hence the rotor currentI 25 becomes,

I 25 =I 15 −Im 5

= 0. 7075 −j 62. 1377 − 0. 0111 +j 2. 272

= 0. 6964 −j 59 .866 amps